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An elementary problem asks for Arg($f(z)$) after a single complete counter-clockwise rotation of the point z about the origin, beginning at the point z = 2 and taking the angle there to be 0, with $f(z) = \sqrt{z^2+2z-3}.$

Plotting a few points by hand and then by computer, I find the final value as Arg(f(z)) = 0. The answer in the text is $\pi.$

I am probably overlooking something obvious...thanks for any insight.

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up vote 2 down vote accepted

Think about $g(z) = \sqrt{z}$. After you go through a complete counter-clockwise rotation, you find that

$$g(z e^{i 2 \pi}) = g(z) e^{i \pi} $$

That is, a $2 \pi$ rotation in $z$ produces a $\pi$ rotation in $\sqrt{z}$. Think about this in light of your function.

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Try it without the linear term and see what you get. – Ron Gordon Jan 10 '13 at 22:14
Yes, so you may conclude that the linear term introduces the $\pi$ phase. This is how a Riemann surface works, by the way. – Ron Gordon Jan 10 '13 at 22:21
Sorry, I was just being suggestive. Maybe I am misreading your last comment, but $z^2+2 z-3 < 0$ when $z \in (-3,1)$. There is a branch cut (another way of looking at the complex plane) in the plane that is this interval. For $|z| = 2$, you should see the $\pi$ phase after a $2 \pi$ rotation. But, if you do not breach the branch, you will not see such a rotation. – Ron Gordon Jan 10 '13 at 22:36
It is $\pi$ because it crosses the branch cut between $z \in (-3,1)$. – Ron Gordon Jan 10 '13 at 22:44
See this: – Ron Gordon Jan 10 '13 at 22:46

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